Question

Does `a_n=x^n/(xn!)` converge for any x?

Answer 1

converges for `x in RR^+`

Explanation:

Using the Stirling asymptotic approximation

`n! approx (n/e)^n` we have

`x^n/((x cdot n)!) approx x^n/((x cdot n)/e)^(x cdot n)=(x/x^x(e/n)^x)^n`

The sequence converges if `x/x^x(e/n)^x<1` so

`e/n < x^((x-1)/x)` Now given a `n` such that `e/n le epsilon`

then the convergence is attained for `x^((x-1)/x) > epsilon`

The function `f(x)=x^((x-1)/x)` for `x in RR^+` has a minimum at `x=1` and `f(1)=1`. Concluding the sequence is convergent for `x in RR^+`

Attached a plot of `f(x)` for `x in RR^+`

Answer 2

Yes, `lim_(n to infty)a_n=0` for all real numbers `x`.

Explanation:

Choose any real number `x`, and let `N` be any natural number greater than or equal to `|x|`.

Let us look at `|a_n|`.

`|a_n|=|x^n/(x n!)|=|x|^(n-1)/(n!) leq (N^(n-1))/(n!)`

`=1/ncdot (N/(n-1)cdot N/(n-2)cdots N/(N+1)) cdot (N/N cdot N/(N-1)cdotsN/1)`

`leq 1/n cdot (N/N cdot N/NcdotsN/N) cdot (N^N/(N!))=1/n(N^N/(N!))`

So, we have

`Rightarrow 0 leq |a_n| leq 1/n(N^N/(N!)) to 0` as `n to 0`

By Squeeze Theorem,

`Rightarrow lim_(n to infty)|a_n|=0`

which implies

`Rightarrow lim_(n to infty)a_n=0`

I hope that this was clear.